My challenge question -- Which side is larger?

G

greg1313

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\(\displaystyle 4^{\tfrac12} \ + \ 4^{\tfrac13} \ + \ 4^{\tfrac14} \ \ \ versus \ \ \ 5\)

or, in other symbols,

\(\displaystyle \sqrt{4} \ + \ \sqrt[3]{4} \ + \ \sqrt[4]{4} \ \ \ versus \ \ \ 5\)


Without using a calculator or computer, please show the steps to determine it. Please hide your solution(s).
 
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greg1313 said:
\(\displaystyle 4^{\tfrac12} \ + \ 4^{\tfrac13} \ + \ 4^{\tfrac14} \ \ \ versus \ \ \ 5\)

or, in other symbols,

\(\displaystyle \sqrt{4} \ + \ \sqrt[3]{4} \ + \ \sqrt[4]{4} \ \ \ versus \ \ \ 5\)


Without using a calculator or computer, please show the steps to determine it. Please hide your solution(s).
Click to expand...
Just in case this is a problem from a math-competition (and you have not disclosed your work/thoughts), I'll post my solution after 6/15/22.
 
lookagain said:
Subhotosh Khan, it is not from a math-competition. You can post immediately.
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Nothing clever - my method is brute force. Using Newton-Raphson iteration I calculated 2^(1/2) and 4^(1/3) correct to 5 decimal places - and added
 
Subhotosh Khan said:
Using Newton-Raphson iteration I calculated 2^(1/2) … correct to 5 decimal places
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You haven't memorized the square root of 2? ?

I've disqualified myself because I'd memorized the 2nd and 3rd (and 4th) roots of 2 a long time ago.

?

[imath]\;[/imath]
 
Otis said:
You haven't memorized the square root of 2? ?

I've disqualified myself because I'd memorized the 2nd and 3rd (and 4th) roots of 2 a long time ago.

?

[imath]\;[/imath]
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I have - up to 3 decimal places. But not 5 decimal places......
 
\(\displaystyle 4^{1/2} + 4^{1/3} + 4^{1/4} \text{ versus } 5\)

\(\displaystyle 2 + 4^{1/3} + \sqrt{2} \text{ versus } 5\)

\(\displaystyle 4^{1/3} \text{ versus } 3 - \sqrt{2}\)

Introduce variable u...

\(\displaystyle (4+u)^{1/3} = 3 - \sqrt{2}\)

If u>0 then the RHS of the original question is larger
If u<0 then the LHS of the original question is larger

Cube both sides :D...

\(\displaystyle 4 + u = (3 - \sqrt{2})^2(3 - \sqrt{2})\)

\(\displaystyle 4 + u = (11 - 6\sqrt{2})(3 - \sqrt{2})\)

\(\displaystyle 4+u = 45 - 29\sqrt{2}\)

\(\displaystyle u = 41 - 29\sqrt{2}\)

Approximate by knowing root 2 to 3dp...

\(\displaystyle u \approx 41 - 29\times1.414\)

\(\displaystyle = 41 - (30\times1.414 - 1.414)\)

\(\displaystyle = 41 - (42.42 - 1.414)\)

\(\displaystyle = 41 - 41.006\)

The above approximation of root 2 is rounded down from the real value, therefore the "41.006" would only become larger if more decimal places are used. Therefore u<0 which implies the LHS of the original question is larger.
 
Cubist said:
\(\displaystyle 4^{1/2} + 4^{1/3} + 4^{1/4} \text{ versus } 5\)

\(\displaystyle 2 + 4^{1/3} + \sqrt{2} \text{ versus } 5\)

\(\displaystyle 4^{1/3} \text{ versus } 3 - \sqrt{2}\)

Introduce variable u...

\(\displaystyle (4+u)^{1/3} = 3 - \sqrt{2}\)

If u>0 then the RHS of the original question is larger
If u<0 then the LHS of the original question is larger

Cube both sides :D...

\(\displaystyle 4 + u = (3 - \sqrt{2})^2(3 - \sqrt{2})\)

\(\displaystyle 4 + u = (11 - 6\sqrt{2})(3 - \sqrt{2})\)

\(\displaystyle 4+u = 45 - 29\sqrt{2}\)

\(\displaystyle u = 41 - 29\sqrt{2}\)

Approximate by knowing root 2 to 3dp...

\(\displaystyle u \approx 41 - 29\times1.414\)

\(\displaystyle = 41 - (30\times1.414 - 1.414)\)

\(\displaystyle = 41 - (42.42 - 1.414)\)

\(\displaystyle = 41 - 41.006\)

The above approximation of root 2 is rounded down from the real value, therefore the "41.006" would only become larger if more decimal places are used. Therefore u<0 which implies the LHS of the original question is larger.
Click to expand...
You don't need to approximate. It just takes an additional two lines.

-Dan
 
topsquark said:
You don't need to approximate. It just takes an additional two lines.
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\(\displaystyle u = 41 - 29\sqrt{2}\)

\(\displaystyle 41-u = 29\sqrt{2}\)

Square...

\(\displaystyle (41-u)^2 = 841\times 2\)

\(\displaystyle u^2 - 82u + 1681 = 1682\)

\(\displaystyle u(u - 82) = 1\) and we can ignore the solution where u>82, which was introduced while squaring, therefore u<0
 
Cubist said:
\(\displaystyle u = 41 - 29\sqrt{2}\)

\(\displaystyle 41-u = 29\sqrt{2}\)

Square...

\(\displaystyle (41-u)^2 = 841\times 2\)

\(\displaystyle u^2 - 82u + 1681 = 1682\)

\(\displaystyle u(u - 82) = 1\) and we can ignore the solution where u>82, which was introduced while squaring, therefore u<0
Click to expand...
Or just ...

[imath]41^2=1681[/imath]
[imath](29\sqrt{2})^2=841\cdot2=1682[/imath]
Therefore [imath]41<29\sqrt{2}[/imath] and [imath]u = 41 - 29\sqrt{2}<0[/imath]
 
This was posted in another forum where my response used the fact that sqrt(2) ~ 1.414 and I was quickly told that I could not use any calculations. Then how the hay can I add up the three values to compare to 5?
 
Steven G said:
This was posted in another forum where my response used the fact that sqrt(2) ~ 1.414 and I was quickly told that I could not use any calculations. Then how the hay can I add up the three values to compare to 5?
Click to expand...

No, the partial phrasing is "I don't recommend doing it your way." That means you do it at your risk, or you would have to be extra careful. For example, how would you know you are getting enough correct decimals?

So, after correcting your typo, you have (approximately 3.414) + 4^(1/3) versus 5, or something similar to that.

What were you going to do after this point?
 
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